Question

Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\sin }^{-1}x$

Answer 1

$\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dsi{n}^{-1}x}$

$\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dsi{n}^{-1}x}\frac{dx}{dx}$

$\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dx}\frac{dx}{dsi{n}^{-1}x}$

$\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dx}=\frac{1}{(\frac{\sqrt{1+x^2}-1}{x}{)}^2+1}\ast \frac{\frac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2}-1}{x^2}$

$⟹\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dx}=\frac{\sqrt{x^2+1}-1}{\left(2\right((x^2+1{)}^{\frac{3}{2}})-x^2-1)}$

$\frac{dsi{n}^{-1}x}{dx}=\frac{1}{\sqrt{1-x^2}}$

Hence we get,$\frac{dta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)}{dx}\frac{dx}{dsi{n}^{-1}x}=\frac{(\sqrt{x^2+1}-1)\left(\sqrt{1-x^2}\right)}{(2\left(\right(x^2+1{)}^{\frac{3}{2}})-x^2-1)}$